最简单的解析几何把题目看错了，第二题求积分，拆 \(\dfrac{1}{(1+sx^{2})(1+x^{2})}\)都拆错了，哎......，偏偏以这种最遗憾的方式结束。
可是没人会为你哀悼，别忘了作业还没写完......
附上答案：初赛试题A卷
考后感想
世界上听起来最难受的一句话就是：
我原本可以做到的......
第16届全国大学生数学竞赛考前复习
一些做题时的details
- 面对高代有关线性变换等式的题目，例如给出:
已知两个线性变换, 满足
这类题目，我们可以考虑将线性变换转换为在对应基的表示矩阵来做，这样一来给出的等式便转换成了矩阵方程，我们可以尝试取迹，看是否能得到与特征值有关的信息，因为我们知道一个矩阵的迹等于其所有特征值的乘积取负。
此外，我们可以尝试对上述矩阵方程左乘或右乘某些矩阵，再取迹，利用迹的性质
面对可交换的线性变换(给出
)，注意回想可交换的线性变换有哪些隐藏性质！ 注意记清楚函数项级数、反常积分、含参量积分连续、可微、可积的条件。
对于给出逐点收敛，要证一致收敛的题目，考虑用有限覆盖定理证明。
对于要放缩的积分，十分注意积分上下限的放缩，比如常见的拆成两部分，一部分趋于
, 一部分趋于一个常数；面对绝对值，更要注意上下限的变换来去掉绝对值；重中之重，面对含三角函数的积分，一定要考虑把积分上下限拆到该三角函数的一个周期中去讨论和放缩！(尤其是面对无穷限的反常积分) 若对一个连续可微函数，给出某点的导数值，注意考虑 Taylor 展开。
对于证明多元函数的一些性质的证明，可先考虑一元的情形，用类似思想去证明多元。
注意基础矩阵
左乘和右乘得到结果的区别： 面对有关矩阵的题目，我们也要注意将矩阵看作一个线性变换，例如一个可逆矩阵可看作一个同构映射；一个上三角矩阵，可看作不变子空间分解；一个分块对角矩阵，可看作线性空间的直和分解，有根子空间直和分解，循环子空间直和分解，还有不变子空间直和分解，三者分别对应Jordan form、frobenius form 和 直接对角化。
关于幂零矩阵，注意它与根子空间和jordan form的联系。且记住
这个恒等式，是根子空间分解理论的根源！ 考虑两个矩阵是否相似，先验证不变量是否相同，例如矩阵的迹和行列式，再考虑其特征多项式；考虑两个矩阵是否合同，先考虑是否为是否为对称阵，再考虑将矩阵写为二次型，并化为规范型，计算其正负惯性指数。
对于给出同时含的原函数和导函数的不等式条件，尝试考虑构造新的函数，利用已知不等式证明构造的函数具有单调性，进而解决问题。对于构造的思路，可以考虑乘指数函数
，注意指数可以多变 (可以是一个抽象函数，也可以是一个具体函数), 但还是具体情况具体分析，这类问题十分多变。 面对一些复杂积分，可以尝试用复变函数理论去计算。
对于给出数列递推式的问题，多去考虑取对数，利用整体法将一个复杂的同项看作一个新的数列，还有就是根据题目形式利用下面两个恒等式估计通项
Fighting！
Closed Sets and Limit Points
Closed Sets and Limit Points
Closed Sets
A subset
EXAMPLE 1 The subset
EXAMPLE 2 In the plane
EXAMPLE 3 In the finite complement
topology on a set
finite complement topology
The finite complement topology (or cofinite topology) on a set
1.The empty set
2.Any subset
This topology is particularly useful when studying properties of
convergence and compactness, as it ensures that every sequence in
EXAMPLE 4 In the discrete topology on the set
EXAMPLE 5 Consider the following subset of the real
line:
Let's talk about the properties of the closed sets.
Theorem 17.1. Let
and are closed. Arbitrary intersections of closed sets are closed.
Finite unions of closed sets are closed.
proof. (1)
and are closed because they are the complements of the open sets and , respectively.
(2) Given a collection of closed sets, we apply DeMorgan's law, Since the sets are open by definition, the right side of this equation represents an arbitrary union of open sets, and is thus open. Therefore, is closed.
(3) Similarly, ifis closed for , consider the equation The set on the right side of this equation is a finite intersection of open sets and is therefore open, Hence is closed.
We should notice the conceptions of "closed set" is relative.
Now when dealing with subspaces, one needs to be careful in using the term "closed set." If
is a subspace of , we say that a set is closed in if is a subset of and if is closed in the subspace topology of (that is, if — is open in )
We have the following theorem:
Theorem 17.2. Let
proof. Assume that
, where is closed in . (See Figure 17.1) Then is open in , so that is open in , by the definition of the subspace topology. But . Hence is open in , so that is closed in . Conversely, assume that is closed in . (See Figure 17.1) Then is open in , so that by definition it equals the intersection of an open set of with . The set is closed in , and , so that equals the intersection of a closed set of with , as desired.
Theorem 17.3. Let
proof. If
is closed in , then equals the intersection of a closed set of with , by the theorem 17.2, so we have . And is also closed in , thus is closed in .
Closure and Interior of a Set
Given a subset
The closure of
The interior of
We shall not make much use of the intenor of a set, but the closure of a set will be quite important.
NOTE:
When dealing with a topological space
We reserve the notation
Theorem 17.4. Let
proof. Let
denote the closure of in . The set is closed in , so is closed in in Theorem 17.2. Since contains , and since by definition equals the intersection of all closed subsets of containing , we must have
On the other hand, we know thatis closed in . Hence by Theorem 17.2, for some set closed in . Then is a closed set containing ; because is the intersection of all such closed sets, we conclude that . Then .
Since the collection of all closed sets in
Theorem 17.5. Let
Then
if and only if every open set containing intersects . Supposing the topology of
is given by a basis, then if and only if every basis element containing intersects .
proof. Consider the statement in (a). It is a statement of the form
. Let us transform each implication to its contrapositive, thereby obtaining the logically equivalent statement (not ) (not ). Written out, it is the following. In this form, our theorem is easy to prove. If is not in , the set is an open set containing that does not intersect , as desired. Conversely, if there exists an open set containing which does not intersect , then is a closed set containing . By definition of the closure , the set must contain , therefore, cannot be in . Statement (b) follows readily if every open set containing intersects , so does every basis element containing , because is an open set. Conversely, if every basis element containing intersects , so does every open set containing , bacause contains a basis element that contains .
Mathematicians often use some special terminology here. They shorten
the statement "
If
EXAMPLE 6 Let
If
EXAMPLE 7 Consider the subspace
Limit Point
There is yet another way of describing the closure of a set, a way that involves the important concept of limit point, which we consider now.
If
EXAMPLE 8 Consider the real line
If
If
Theorem 17.6. Let
proof. If
in , every neighborhood of intersects (in a point different from x). Therefore, by Theorem 17.5, belong to . Hence . Since by definition , it follows that .
To demonstrate the reverse inclusion, we letbe a point of and show that . If happens to lie in , it is trivial that ; suppose that does not lie in . Since , we know that every neighborhood of intersects ; because , the set must intersect in a point different from . Then , so that , as desired.
Corollary 17.7 A subset of a topological space is closed if and only if it contains all its limit points.
proof. The set
is closed , and the latter holds .
Hausdorff Spaces
One's expenence with open and closed sets and limit points in the real line and the plane can be misleading when one considers more general topological spaces.
For example, in the spaces
and , each one-point set is closed. This fact is easily proved; every point different from has a neighborhood not intersecting , so that is its own closure.
But this fact is not true for arbitrary topological spaces. Consider the topology on the three-point setindicated in Figure 17.3. In this space, the one-point set is not closed, for its complement ( ) is not open.
And, in topological space
Similarly, one's experience with the properties of convergent sequences in
and can be misleading when one deals with more general topological spaces. In an arbitrary topological space, one says that a sequence of points of the space to the point of provided that, corresponding to each neighborhood of , there is a positive integer such that for all . In and , a sequence cannot converge to more than one point, but in an arbitrary space, it can. In the space indicated in Figure 17.3, for example, the sequence defined by setting for all converges not only to the point , but also to the point and to the point !
Topologies in which one-point sets are not closed, or in which sequences can converge to more than one point, are considered by many mathematicians to be somewhat strange. They are not really very interesting, for they seldom occur in other branches of mathematics.
Therefore, one often imposes an additional condition that will rule out examples like this one, bringing the class of spaces under consideration closer to those to which one's geometric intuition applies. The condition was suggested by the mathematician Felix Hausdorff, so mathematicians have come to call it by his name.
Definition. A topological space
Theorem 17.8. Every finite point set in a Hausdorff
space
proof. It suffices to show that every one-point set
is closed. If is a point of different from , then and have disjoint neighborhoods and , respectively. Since does not intersect , the point cannot belong to the closure of set . As a result, the closure of the set is itself, so that is is closed.
The condition that finite point sets be closed is in fact weaker than the Hausdorff condition.
For example, the real line
in the finite complement topology is note a Hausdorff space, but it is a space in which finite point sets are closed.
The condition that finite point sets be closed has been given a name
of its own: it is called the
Theorem 17.9. Let
proof. If every neighborhood of
intersects in infinitely many points, it certainly intersects in some point other than itself, so that is a limit point of .
Conversely, supposeis a limit point of , and suppose some neighborhood of intersects in only finitely many points. Then also intersects in finitely many points; let be the points of . The set is an open set of , since the finite point set is closed; then is a neighborhood of x that intersects the set not at all. This contradicts the assumption that is a limit point of .
One reason for our lack of interest in the
Theorem 17.10. If
proof. Suppose that
is a sequence of points of that converges to . If , let and be disjoint neighborhoods of and , respectively. Since contains for all but finitely many values of , the set cannot. Therefore, cannot converge to .
If the sequence
Theorem 17.11.
(1). Every simply ordered set is a Hausdorff space in the order topology.
(2). The product of two Hausdor-if spaces is a Hausdorff space.
(3). A subspace of a Hausdorff space is a Hausdorff space.
proof.
(1). Given two elementsof the simply orderd set , and . We can find an interval be the neighborhood of , and be the neighborhood of such that . Thus, , is a Hausdorff space.
(2). Suppose thatand are the two Hausdorff spaces, and is the product topology. Given , then . We can find a neighborhood of , and of such that , we also can find a neighborhood of , and of such that . Hence , and . So is a Hausdorff space. (3). Suppose is a Hausdorff space, . Obviously, is also a Hausdorff space.
The order Topology, Product Topology, Subspace Topology
Order Topology
If
Suppose that
The notation used here is familiar to you already in the case where X is the real line, but these are intervals in an arbitrary ordered set. A set of the first type is called an open interval in X, a set of the last type is called a closed interval in X, and sets of the second and third types are called half-open intervals
Definition. Let
All open intervals
in . All intervals of the form
, where is the smallest element (if any) of . All intervals of the form
, where is the largest element (if any) of .
The collection
if
Note: The elements of the order topology are not the intervals, we should notice the definition of the interval!!!
We need to check that
First, note that every element
of lies in at least one element of : The smallest element (if any) lies in all sets of type (2), the largest element (if any) lies in all sets of type (3), and every other element lies in a set of type (1).
Second, note that the intersection of any two sets of the preceding types is again a set of one of these types, or is empty.
EXAMPLE 1 The standard topology on
EXAMPLE 2 Consider the set
EXAMPLE 3 The positive integers
EXAMPLE 4 The set
Rays
Definition. If
The use of the term "open" suggests that open rays in X are open sets in the order topology. And so they are.
Consider, for example, the ray
. If has a largest element , then equals the basis element . If has no largest element, then equals the union of all basis elements of the form , for . In either case, is open. A similar argument applies to the ray .
The open rays, in fact, form a subbasis for the order topology on
Because the open rays are open in the order topology, the topology they generate is contained in the order topology.
On the other hand, every basis element for the order topology equals a finite intersection of open rays; the intervalequals the intersection of and , while and , if they exist, are themselves open rays. Hence the topology generated by the open rays contains the order topology.
The Product Topology on
If
Definition. Let
Let's check that
The first condition is trivial, since
is itself a basis element. The second condition is almost as easy, since the intersection of any two basis elements and is another basis element. For and the latter set is a basis element because and are open in and , respectively. See the figure below, it can help you understand it intutively.
Note: The collection
We may ask what's the basis of the topology of
Theorem 15.1. If
proof. We apply Lemma 13.2. Given an open set
of and a point of , by definition of the product topology there is a basis element such that . Because and are bases for and , respectively, we can choose an element such that , and an element such that . Then . Thus the collection meets the criterion of Lemma 13.2, so is a basis for .
EXAMPLE 1 We have a standard topology on
It is sometimes useful to express the product topology in terms of a subbasis. To do this, we first define certain functions called projections.
Definition. Let
We use word "onto" because
If
Theorem 15.2. The collection
proof. Let
denote the product topology on . let be the topology generated by . Because every element of belongs to , so do the arbitrary unions of finite intersections of elements of . Thus . On the other hand, every basis element for the topology of is a finite intersection of elements of , since Therefore, , so that , hence , as desired.
The subspace Topology
Definition. Let
It is easy to see that
It contains
and because where and are elements of .
The fact that it is closed under finite intersections and arbitrary unions follows from the equations
Lemma 16.1. If
proof. Given
open in and given , we can choose an element such that . Then . It follows from Lemma 13.2 that is a basis for the subspace topology on .
When dealing with a space
We make the following definition If
There is a special situation in which every set open in
Lemma 16.2. Let
proof. Since
is open in , for some set open . Since and are both open in , so is .
Now let us explore the relation between the subspace topology and the order and product topologies.
Theorem 16.3. If
proof.The set
is the general basis element for , where is open in and is open in . Therefore, is the general basis element for the subspace topology on . Now Since and are the general open sets fot the subspace topologies on the and , respectively, the set is the general basis element for the product topology on .
Thus, we know that the bases for the subspace topology onand for the product topology on are the same. Hence the topologies are the same.
Now let
However, the resulting order topology on
EXAMPLE 1 Consider the subset
Note that these sets form a basis for the order topology on
Y. Thus, we see that in the case of the set
EXAMPLE 2 Let
EXAMPLE 3 Let
For example, the set
is open in in the subspace topology, but not in the order topology, because we choose , any basis element for the order topology on that contains is of the form such a set is necessarily not be contained in
The set
Convex
Given an ordered set
Theorem 16.4 Let
proof. Consider the ray
in . What is its intersection with ? If , then this is an open ray of the orderd set . If , then is either a lower boud on or an upper bound on , since is convex. In the former case, the set equals all of ; in the latter case, is is empty.
A similar work remark show that the intersection of the raywith is either an open ray of , or itself, or empty. Since the sets and form a subbasis for the subspace topology on , and since each is open in the order topology, the order topology contains the subspace topology.
To prove the reverse, note that any open ray of Y equals the intersection of an open ray ofwith , so it is open in the subspace topology on . Since the open rays of are a subbasis for the order topology on , this topology is contained in the subspace topology, hence they are the same.
Comments: In this proof, we need to be familiar with the definition of rays and its properties (such that it is the subbasis for the order topology)
To avoid ambiguity, let us agree that whenever
Measure theory
The exterior measure
Definition. If
Info Note
It is important to note that it would not suffice to allow finite sumsin the definition of
One can, however, replace the coverings by cubes, with covering by rectangles; or with coverings by balls. That the former alternative yields the same exterior measure is quite direct.
EXAMPLE 1. The exterior measure of a point is zero. The exterior measure of the empty set is also zero.
This is clear that a point is a cube with volume zero, and which covers itself, so does the empty set.
EXAMPLE 2. The exterior measure of a closed cube is equal to its volume.
Suppose
is a closed cube in . Since covers itself, we have .Therefore, it suffices to prove the reverse inequality.
We consider an abitraty coveringby cubes, and by the definition of the exterior measure, it suffices to prove that For a fixed we choose for each an open cube which contains , and such that . From the open covering of the compact set , we may select a finite subcovering, which after possibly renumbering the rectangles, we may write as . Taking the closure of the cubes , we have . Consequently, Since is arbitrary, we find the inequality before we need to prove holds; thus , as desired.
EXAMPLE 3. If
Since
is covered by its closure , and , we immediately see that , and obviously by the definition of exterior measure.
EXAMPLE 4. The exterior measure of a rectangle
Arguing as in Example 2, we see that
. To obtain the reverse inequality, consider a grid in formed by cubes of sides length . Then, if consists of the (finite) collection of all cubes entirely contained in , and the (finite) collection of all cubes that intersect the complement of , we first note that . Also, a simple argument yields Moreover, there are cubes in , and these cubes have volume , so that . Hence and letting tend to infinity yields , as desired.
EXAMPLE 5. The exterior measure of
This follows from the fact that any covering of
is also a covering of any cube , hence . Since can have arbitrarily large volume, we must have
EXAMPLE 6. The Cantor set
From the construction of
, we know that , where each is a disjoint union of closed intervals, each of length . Consequently, for all , hence .
Observation 1 (Monotonicity) If
This follows that any coverings of
by a countable collection of cubes is also a covering of .
In particular, monotonicity implies that every bounded subset of
Observation 2 (Countable sub-additivity) If
First, we may assume that each
, for otherwise the inequality clearly holds. For any , the definition of exterior measure yields for each a covering by closed cubes with Then, is a covering of by closed cubes, and therefore Since this holds true for every , the second observation is proved.
Observation 3 If
By monotonicity, it is clear that the inequality
holds. For the reverse inequality, let and choose cubes such that , with Let denote an open cube containing , and such taht . Then is open, and by Observation 2 Hence , as was shown.
Observation 4 If
By Observation 2, we already know that
, so it is suffices to prove the reverse inequality. To this end, we first select such that . Next, we choose a covering by closed cubes, with . We may, after subdividing the cubes , assume that each has a diameter less than . In this case, each can intersect at most one of the two sets or . If we denote by and the sets of those indices for which intersects and , respectively, then is empty, and we have Therefore
Observation 5 If a set
Let
denote a cube strictly contained in such that , where is a arbitrary but fixed. Then, for every , the cubes are disjoint, hence at a finite distance from one another, and repeated applications of Observation 4 imply Since , we conclude that for every integer , When the tend to infinity, we deduce for every , hence , as desired.
Info Note
This last property shows that if a set can be decomposed into almost disjoint cubes, its exterior measure equals the sum of the volumes of the cubes. Moreover, this also yields a proof that the sum is independent of the decomposition
Measureable sets and the Lebesgue measure
Definition. A subset
If
Let's study the properties of the Lebesgue measure.
Property 1 Every open set in
Property 2 If
By Observation 3 of the exterior measure, for every
there exists an open set with and . Since , monotonicity implies , as desired.
Property 3 A countable union of measurable sets is measurable.
Suppose
, where each is measurable. Given , we may choose for each an open set with and . Then the union is open, , nd , so monotonicity and sub-additivity of the exterior measure imply
Property 4 Closed sets are measurable
First, we observe that it suffices to prove that compact sets are measurable. Indeed, any closed set
can be written as the union of compact sets, say , where denotes the closed ball of radius centered at the origin; then Property 3 applies.
So supposeis compact (so that in particular ), and let . Since is closed, the difference is open, and by Theorem 1.4 we may write this difference as a countable union of almost disjoint cubes FOr a fixed , the finite union is compact; therefore (we isolate this little fact in a lemma below). Since , Observation 1, 4, and 5 of the exterior measure imply Hence , and this also holds in the limit as tends to infinity. Invoking the sub-additivity property of the exterior measure finally yields as desired.
Theorem 1.4. Every open subset
Lemma 3.1. If
proof. Since
is closed, for each point , there exists so that . Since covers , and is compact, we may find a subcover, which we denote by . If we let , then we must have . Indeed, if and , then for some we have , and by construction . Therefore and the lemma is proved.
Compact set: A bounded set is compact if it is also closed. Compact sets enjoy the Heine-Borel covering property.
Property 5 The complement of a measurable set is measurable.
If E is measurable, then for every positive integer
we may choose an open set with and . The complement is closed, hence is measurable, which implies that the union is also measurable by Property 3. Now we simply note that , and such that for all . Therefore, , and is measurable by Property 2. Therefore is measurable since it is the union of two measurable sets, namely and .
Property 6 A countable intersection of measurable sets is measurable.
This follows from Properties 3 and 5, since
Warning
we have proved that the collection of measurable sets is closed under countable unions and intersections. We emphasize, however, that the operations of uncountable unions or intersections are not permissible when dealing with measurable sets!
Theorem 3.2. If
proof. First, we assume further that each
is bounded. Then, for each , by applying the definition of measurability to , we can choose a closed subset or with . For each fixed , the set are compact and disjoint, so that . Since , we must have Letting tend to infinity, since was arbitrary we find that Since the reverse inequality always holds(by sub-aditivity in Observation 2), this concludes the proof when each is bounded.
In the general case, we select any sequence of cubesthat increases to , in the sense that for all and . We then let and for . If we define measurable sets by . then The union above is disjoint and every is bounded. Moreover , and this union is also disjoint. Putting these facts together, and using what has already been proved, we obtain as claimed.
We make two definitions to state succinctly some further consequences.
If
Similarly, if
Corollary 3.3 Suppose
If
, then . If
and for some k, then
proof. For the
first part, let
>For the second part, we may clearly assume that
Info Note
The reader should note that the second conclusion may fail without
the assumption that
What follows provides an important geometric and analytic insight into the nature of measurable sets, in terms of their relation to open and closed sets.
an arbitrary measurable set can be well approximated by the open sets that contain it, and alternatively, by the closed sets it contains.
Theorem 3.4 Suppose
There exists an open set
with and . There exists a closed set
with and . If
is finite, there exists a compact set with and . If
is finite, there exists a finite union of closed cubes such that The notation stands for the symmetric difference between the sets and , defined by , which consists of those points that belong to only one of the two sets or .
proof. Part(i) is just the definition of the measurability.
For the second part, we know that
is measurable, so there exists an open set with and . If we let , then is closed, , and . Hence as desired. Comments: If we meet the closed set, we can consider its complement, because its complement is open set.
For (iii), we first pick a closed set
so that and . For each , we let denote the ball centered at the origin of radius and define compact sets . Then is a sequence of measurable that decreases to , and since , we conclude that all large one has . For (iv), choose a family of closed cubes
so that Since , the series converges and exists such that . If , then
Invariance properties of Lebesgue measure
Translation-invariance: If
To prove the measurability of
under the assumption that is measurable, we note if is open, , and , then is open, , and
Relative dilation-invariance: Suppose
Reflection-invariant: Whenever
Basis for a Topology
Basis for a Topology
Definiton. If
- For each
, there is at least one basis element containing - If
, then there is a basis element containing such that
If
Note: Every basis element is itself an element of
Example 1. Let
Example 2. Let
Example 3. If
We need to check that the collection
Let us check now that the collection
If
is the empty set, it satisfies the defining condition of openness vacuously. Now let us take an indexed family of elements of and show that Given , there is an index such that . Since is open, there is an basis element such that , then , so that is open, by definition.
Now let us take two elementsand of and show that . Given , choose a basis element containing such that , and containing such that . Hence , and by the second condition for a basis, there is a basis element such that . Then , so , by definition.
Finally, we show by induction that any finite intersectionof elements of is in . This fact is trivial for ; we suppose it true for and prove it for . Now By hypothesis, ; by the result just proved the intersection of and also belong to .
Thus we have checked that collection of open sets generated by a basisis, in fact, a topology.
Another way of describing the topology generated by a basis is given in the following lemma:
Lemma 13.1. Let
proof. we need to prove any subset
of satisfies the equation First given a collection of elements of ,there are also elements of . Because is a topology, their union is in . Conversely, given . Choose each , there exists an element of such that , thus , and it is obvious that , Thus, .
This lemma states that every open set
We have described in two different ways how to go from a basis to the topology it generates. Sometimes we need to go in the reverse direction, from a topology to a basis generating it. Here is one way of obtaining a basis for a given topology; we shall use it frequently.
Lemma 13.2. Let
proof. We must show that
is a basis, thus we need to check the two conditions of definition of basis, given , since is itself an open set, there is by hypothesis an element of such that . To check the second condition, let , where and are elements of . Since and are open, so is , then there exists by hypohesis an element such that .
Letbe a collection of open sets of ; we must show that the topology generated by equals the topology . First, note that if and if , then there is by hypohesis an element of such that . If follows that , by definition. Conversely, if , then by the preceding lemma. Since each element of belongs to and is a topology, also belong to .
We can use the bases to determine whether one topology is finer than another.
Lemma 13.3. Let
is finer than . For each
and each basis element containing , there is a basis element such that .
proof. (2)
(1). Given a element of , we wish to show that . Let , since generate , there is a basis element such that . Condition (2) tell us that there exist an element such that , then , so , by definition.
(1)(2). Given a element and , then , and by condition (1), we have , thus . Since is generated by , there is an element such that .
Example 4. We can see that the collection
We now define three topologies on the real line
Definition. If
It is easy to see that all three of these collections are bases; in each case, the intersection of two basis elements is either another basis element or is empty. The relation between these topologies is the following.
Lemma 13.4. The topologies of
proof. Let
be the topologies of , respectively. Given a basis element for and a point , there exists a basis element such that , thus is finer than by the Lemma 13.3.
A similar argument applies to. Given a basis element for and a point , there exist a basis element that contains . On the other hand, given a basis element for and point of , there is no open interval contains and lies in .
Subbasis
Definition. A subbasis
We must of course check that
Above all,
吐槽hexo渲染数学公式
hexo渲染latex编写的数学公式真是一言难尽啊，给我整了好久，之前说什么latex的"|"会被nunjucks模板误解，导致编译错误，然后我先去掉出错的代码，运行
接着，直接重新把之前出错的代码再粘贴回去，刷新服务器，我发现原来出错的代码可以正确显示公式，现在我再运行
1
$ hexo s
又发现它又不报错了，无语......1
2$ hexo g
$ hexo d
数学类-2020-A卷-T6
对于有界区间
(1).
Hello World
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